If the ratio of roots of $ax^2+bx+c=0$ be equal to that of $a'x^2+b'x+c'=0$ then If the ratio of roots of $ax^2+bx+c=0$ be equal to that of $a'x^2+b'x+c'=0$ then prove that:
$$\dfrac {b^2}{b'^2} = \dfrac {ac}{a'c'}$$
My Attempt:
Let $\alpha $ and $\beta $ be the roots of the first equation and $\alpha' $ and $\beta' $ be the roots of the second equation. Then,
$$\alpha + \beta = \dfrac {-b}{a}$$
$$\alpha. \beta = \dfrac {c}{a}$$
$$\alpha' + \beta'=\dfrac {-b'}{a'}$$
$$\alpha'.\beta'=\dfrac {c'}{a'}$$
According to Question:
$$\dfrac {\alpha }{\beta}=\dfrac {\alpha' }{\beta'}$$
 A: Let $$\frac{\alpha}{\beta} = \frac{\alpha'}{\beta'} = k.$$
Now, $\alpha+\beta = -\frac{b}{a} \implies (k+1)a\beta=-b$. Similarly,  $(k+1)a'\beta'=-b'$. Therefore,
$$\frac{b^2}{b'^2}=\frac{a(a\beta^2)}{a'(a'\beta^2)}.$$
But $\alpha \beta = \frac{c}{a} \implies c = ka\beta^2$. Similarly, $c' = ka'\beta'^2$. Therefore, $$\frac{c}{c'}=\frac{a \beta^2}{a' \beta'^2} \implies \frac{b^2}{b'^2}=\frac{ac}{a'c'}.$$
A: I would say: 
$\frac {\alpha'}{\alpha'} = \frac {\alpha}{\beta}$ implies that there exists a constant $k$ such that $\alpha' = k\alpha, \beta' = k\beta$
$\frac {b'}{a'} = -(\alpha' + \beta') = -k(\alpha + \beta) = k\frac {b}{a}$
$\frac {c'}{a'} = \alpha'\beta' = k^2\alpha\beta = k^2\frac {c}{a}$
$\frac {a'b}{ab'} = k$
$\frac {a'^2b^2}{a^2b'^2} = k^2 = \frac {a'c}{ac'}$
$\frac {b^2}{b'^2}= \frac {ac}{a'c'}$
A: Using your identities:
\begin{align}
\frac{b^2}{b'^2}&=\frac{a^2(\alpha+\beta)^2}{a'^2(\alpha'+\beta')^2}\\
&=\frac{a^2\alpha\left(1+\frac{\beta}{\alpha}\right)\beta\left(\frac{\alpha}{\beta}+1\right)}{a'^2\alpha'\left(1+\frac{\beta'}{\alpha'}\right)\beta'\left(\frac{\alpha'}{\beta'}+1\right)}\\
&=\frac{a^2\alpha\beta}{a'^2\alpha'\beta'}\\
&=\frac{a^2\frac{c}{a}}{a'^2\frac{c'}{a'}}\\
&=\frac{ac}{a'c'}
\end{align}
